Assignment 21 + t-test Practice

We talked about this assignment 21 a lot on Thursday, so I don't feel the need to go over it a whole lot. We answered all of the "Plan" section, and even one of the answers to the "Plan" section of part B.

A common mistake on this one is the "list and state how the conditions are met". This means you must STATE the condition (like Normality:) then after the colon, state how it is met for this particular problem.

Remember how the conditions change slightly for a t-test.

On part B, be careful with the t-test confidence interval: we are using t*, not Z*, which I think is a common mistake.

Okay, now onto t-test practice so you can actually solve the problems!

Let's do an example!

Hillary thinks that the statistic department isn't correctly stating the actual amount of late-fee money they receive from Stat 121 students. They claim that on average each test gives them 5,000 dollars. Hillary takes a simple random sample of a 10 different testing periods over the last five years and gets a mean of 5,800 dollars and standard deviation of 750. Alpha = 0.05. Assume test fees are normally distributed.

STATE: Is the true mean income earned by Stat 121 late fees greater than 5,000 dollars?

Okay. So there are a few things we notice here off the bat. First where is the standard deviation from? It says in the problem it is from the sample, meaning that we know S, not sigma. This means we will be doing a t-test. Also, the STATE lets us know what our hypothesis will end up being (greater than).

For the sake of this problem, we aren't going to go through the entire Plan or Solve steps, only because the point of this problem is to help you learn how to use the t-table.

Ho: Mu=5000
Ha: Mu > 5000

t=5800-5000/ [750/sqrt(10)] = 3.37

Now we go to the t chart. We need one more thing though before we use it: degrees of freedom. Remember, df= n-1.

So in this case, df= 10-1 = 9.

Go to the tenth row in the t-table. Find the two values that sandwich our t value.

I see that the t* values of 3.690 and 4.397.

I then trace my fingers down to the "one sided t test" row (because we are greater than) and read off the two p value values: .005 and .0025

Thus I can say my p value is: .0025< p value < .oo5.

Conclude as usual.

Now let's try a t confidence interval.

The only thing that changes for a t test versus a z test is we are finding t* instead of z*.

Let's try finding t* using our problem above.

We need two things to find t*. One, degrees of freedom which we already found to be 9. Second is confidence level. Since we had an alpha= 0.05, it follows that our confidence level is 95%.

Now we simply find where our df and confidence level intersect. This is our t*.

From the table, I get t*= 2.262.

Hope that helps! Remember don't wait to do assignment 21. The open lab will be overflowing on Wednesday. Get it done! As always if you have questions email me!

-Hillary